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anonymous

6 years ago

I understand the process of rotating a region with the x-axis and other lines parallel to it, but I keep having trouble formulating the integral of the equation representing the volume whenever the rotation is in reference with the y-axis and its meridians.

anonymous

6 years ago

What I usually do in such case is that I integrate the function with respect to y

anonymous

6 years ago

The region is bounded by the curves of:\[y=x ^{2/3}\]\[y=0\]\[x=0\]The rotation is in reference with the y-axis

|dw:1333616173853:dw| This a rough sketch of how the graphs should look

anonymous

6 years ago

excuse be, may I ask is it really y=0 and x=0?

anonymous

6 years ago

So my parameters for the integral would be the min and max of the intersection?

anonymous

6 years ago

I'm sorry, the rotation is in reference with the line x=0 but the region is above the line y=0.

TuringTest

6 years ago

|dw:1333616352713:dw|can't be just those three bounds, what's the fourth?
there is no intersection besides at the origin, unless you have omitted an eqn.

TuringTest

6 years ago

what about on the right side though?

anonymous

6 years ago

I did once again I'm sorry the line I drew in the first drawing should be the line x=1 and the intersection is at (1,1).

anonymous

6 years ago

ah ok

TuringTest

6 years ago

|dw:1333616595171:dw|now which axis are we going around?

anonymous

6 years ago

so the cross sectional area A is pie(x^(2/3))^2

anonymous

6 years ago

I understand that I need to rewrite all the equations in terms of x, that is having the x as dependent on the y var.

anonymous

6 years ago

then we know we'll integrate from 0 to 1 so
\[\int\limits_{0}^{1}\pi(x^{2/3})^2dx\]

anonymous

6 years ago

The axis would be the y-axis.

anonymous

6 years ago

that is if we are to rotate about the x axis

anonymous

6 years ago

Don't I need to subtract that volume by the volume of the cylinder formed by the line x=1 with radius 1?

anonymous

6 years ago

yes

TuringTest

6 years ago

that is one way
the other is shell method

TuringTest

6 years ago

...but then you would integrate along y

anonymous

6 years ago

@TuringTest hmmm.. can you tell something about this?

TuringTest

6 years ago

oh no, srry, thinking of going around x-axis again

anonymous

6 years ago

I think I'm getting it after all, what I had to do since the very beginning was change the equation y=x^(2/3) to x=y^(3/2) and use that as the radius for the inner disk and 1 as the radius for the outer disk.

TuringTest

6 years ago

yeah that would work fine

TuringTest

6 years ago

actually, that's probably the easiest way
as I said last time, more than one way to skin a cat

anonymous

6 years ago

and from there proceed to integrate from 0 to to 1 since I'm 'slicing' this figure in horizontal sections.

anonymous

6 years ago

Yeah, I'm having a test over this tomorrow and I need to know this by heart.

TuringTest

6 years ago

well, vertical sections|dw:1333617106925:dw|here's a visual for you of two of the rings you are talking about
(the best I can do with what I'm given)

TuringTest

6 years ago

the inner radius is as you said: \(r_i=x=y^{3/2}\) and the outer is\(r_o=1\)
the intesrsection is still \((1,1)\)
so you can look at it as subtracting from the cylinder of radius 1 if you wish...

TuringTest

6 years ago

|dw:1333617404061:dw|

anonymous

6 years ago

Yes I meant to say that I was slicing this 'cylinder' from left to right. Thanks for the clarification.

TuringTest

6 years ago

I think you'll do fine on your test
I'm tired... going to sleep
good night :)